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Chemistry · EQUILIBRIUM

हिन्दी

For the dissociation of gaseous hydrogen iodide, \( 2\text{HI}(\text{g}) \rightleftharpoons \text{H}_2(\text{g}) + \text{I}_2(\text{g}) \), if the degree of dissociation is \( \alpha \), the equilibrium constant \( K_c \) in terms of \( \alpha \) is:

  1. A
    \( \frac{4\alpha^2}{(1-\alpha)^2} \)
  2. B
    \( \frac{\alpha^2}{(1-\alpha)^2} \)
  3. C
    \( \frac{\alpha^2}{4(1-\alpha)^2} \)
  4. D
    \( \frac{\alpha^2}{2(1-\alpha)} \)
View Answer & Solution

Correct answer · Option C

\( \frac{\alpha^2}{4(1-\alpha)^2} \)

Explanation

Starting with \( C\text{ mol L}^{-1} \) of \( \text{HI} \): at equilibrium, \( [\text{HI}] = C(1-\alpha) \), \( [\text{H}_2] = \frac{C\alpha}{2} \), \( [\text{I}_2] = \frac{C\alpha}{2} \). Thus, \( K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} = \frac{(C\alpha/2)^2}{[C(1-\alpha)]^2} = \frac{\alpha^2}{4(1-\alpha)^2} \).
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